\(A\left(x\right)+B\left(x\right)=8x^3-4x-6+2x^3-x^2-3x+6=10x^3-x^2-7x\)

Hàng ngang:

A(x) +B(x)=5x³+3x³-4x-6+2x³-x²-3x+6

                 = 10x³-x²-7x

Hàng dọc:

A(x)=5x³+3x³-4x-6=8x³-4x-6

   8x³      -4x-6

+ 2x³-x²-3x+6

----------------------

   10x³-x²-7x

       x³  -2x² + x -1

    x

                  -x+ 5

     5x³ -10x²+5x-5

-x⁴+ 2x³- x² +x  

-x⁴+7x³-11x²+6x-5

sao không àm ngang được à

(x³-2x²+x-1)(5-x)

= 5.x³-2x².5+x.5-1.5 -x.x³ -2x².x +x.x-1.x

=5x³ -10x²+ 5x -5-x⁴ -2x³+x² -x

=-x⁴ +3x³ -9x² +4x

ai giúp mk với nè , tối học rồi

b: A(x)*C(x)

=(x^3-3x^2+5x-3)(x-3)

=x^4-3x^3-3x^3+9x^2+5x^2-15x-3x+9

=x^4-6x^3+14x^2-18x+9

c: B(x):C(x)

\(=\dfrac{-x^3+3x^2-x^2+3x-6x+18-13}{x-3}\)

=-x^3-x-6-13/x-3

d: A(x)-B(x)

=x^3-3x^2+5x-3+x^3-2x^2+3x-5

=2x^3-5x^2+8x-8

a: =3x^3+12x-9-2x^4-8x^2+6x

=-2x^4+3x^3-8x^2+18x-9

b: \(=\dfrac{-2x^4+x^3+6x^3-3x^2}{2x+1}=-x^3+3x^2\)

a) \(=x^4-14x^2+40-72=x^4-14x^2-32=\left(x-4\right)\left(x+4\right)\left(x^2+2\right)\)

b) \(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)+1=\left(x^2+5x\right)^2+2\left(x^2+5x\right)+1=\left(x^2+5x+1\right)^2\)

c) \(=x^4+3x^3-3x^2+3x^3+9x^2-9x+x^2+3x-3-5=x^4+6x^3+7x^2-6x-8=\left(x-1\right)\left(x+1\right)\left(x+2\right)\left(x+4\right)\)

a: Ta có: \(\left(x^2-4\right)\left(x^2-10\right)-72\)

\(=x^4-14x^2-32\)

\(=\left(x^2-16\right)\left(x^2+2\right)\)

\(=\left(x-4\right)\left(x+4\right)\left(x^2+2\right)\)

b: Ta có: \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)+1\)

\(=\left(x^2+5x+6\right)\left(x^2+5x+4\right)+1\)

\(=\left(x^2+5x\right)^2+10\left(x^2+5x\right)+24+1\)

\(=\left(x^2+5x+1\right)^2\)